proving $ \lim_{x\to \infty}$ ${x+3\over x-2}$=1 using delta epsilon method

Question

I need to prove $ \lim_{x\to \infty}$ ${x+3\over x-2}$=1 using delta-epsilon method. Here's the part I tried but I'm stuck in the middle in choosing M with $\varepsilon$. Let $\varepsilon$ >0 be any number < we need to find a number x>M whenever $\vert {x+3\over x-2} -1 \vert $ < $\varepsilon$. From this, we can get $\vert {5\over x-2} \vert $ < $\varepsilon$. And | x-2| > ${5 \over \varepsilon}$. So after this should I take M= ${5 \over \varepsilon}$ and continue?I don't get how to compare M<x with this. Please can someone help with solving this?

Answer 1

$ | \frac{5}{x-2} | < \epsilon \iff| \frac{1}{x-2} | < \frac{\epsilon }{5} \iff | x-2 | > \frac{5}{\epsilon}\ \iff (x> 5/\epsilon + 2) or ( x < 2 - 5/\epsilon) $. So, all you have to do is put $ M = \frac{5}{\epsilon} + 2$, and by the above argument, the definition of the limit will be satisfied.